Question

determine a value of the constant a for which $lim_{x
ightarrow - 1}g(x)$ exists and state the value of the limit, if possible.
g(x)=$\begin{cases}x^{2}-2x & \text{if }xleq - 1\ax^{3}-5 & \text{if }x > - 1end{cases}$
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the limit exists for a = . with that value for a, $lim_{x
ightarrow - 1}g(x)$ = .
b. there is no value of a for which $lim_{x
ightarrow - 1}g(x)$ exists.

Explanation:

Step1: Find the left - hand limit

The left - hand limit as $x\to - 1$ is when $x\leq - 1$. So, $\lim_{x\to - 1^{-}}g(x)=\lim_{x\to - 1^{-}}(x^{2}-2x)$. Substitute $x = - 1$ into $x^{2}-2x$: $(-1)^{2}-2\times(-1)=1 + 2=3$.

Step2: Find the right - hand limit

The right - hand limit as $x\to - 1$ is when $x> - 1$. So, $\lim_{x\to - 1^{+}}g(x)=\lim_{x\to - 1^{+}}(ax^{3}-5)$. Substitute $x=-1$ into $ax^{3}-5$: $a\times(-1)^{3}-5=-a - 5$.

Step3: Set the left - hand and right - hand limits equal

For the limit $\lim_{x\to - 1}g(x)$ to exist, $\lim_{x\to - 1^{-}}g(x)=\lim_{x\to - 1^{+}}g(x)$. So, $3=-a - 5$.

Step4: Solve for a

Add 5 to both sides of the equation $3=-a - 5$: $3 + 5=-a$, i.e., $8=-a$. Then multiply both sides by - 1 to get $a=-8$.

Step5: Find the limit value

When $a = - 8$, the limit $\lim_{x\to - 1}g(x)=3$.

Answer:

A. The limit exists for $a=-8$. With that value for $a$, $\lim_{x\to - 1}g(x)=3$